-0.000 254 797 590 553 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 553(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 254 797 590 553(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 254 797 590 553| = 0.000 254 797 590 553


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 254 797 590 553.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 254 797 590 553 × 2 = 0 + 0.000 509 595 181 106;
  • 2) 0.000 509 595 181 106 × 2 = 0 + 0.001 019 190 362 212;
  • 3) 0.001 019 190 362 212 × 2 = 0 + 0.002 038 380 724 424;
  • 4) 0.002 038 380 724 424 × 2 = 0 + 0.004 076 761 448 848;
  • 5) 0.004 076 761 448 848 × 2 = 0 + 0.008 153 522 897 696;
  • 6) 0.008 153 522 897 696 × 2 = 0 + 0.016 307 045 795 392;
  • 7) 0.016 307 045 795 392 × 2 = 0 + 0.032 614 091 590 784;
  • 8) 0.032 614 091 590 784 × 2 = 0 + 0.065 228 183 181 568;
  • 9) 0.065 228 183 181 568 × 2 = 0 + 0.130 456 366 363 136;
  • 10) 0.130 456 366 363 136 × 2 = 0 + 0.260 912 732 726 272;
  • 11) 0.260 912 732 726 272 × 2 = 0 + 0.521 825 465 452 544;
  • 12) 0.521 825 465 452 544 × 2 = 1 + 0.043 650 930 905 088;
  • 13) 0.043 650 930 905 088 × 2 = 0 + 0.087 301 861 810 176;
  • 14) 0.087 301 861 810 176 × 2 = 0 + 0.174 603 723 620 352;
  • 15) 0.174 603 723 620 352 × 2 = 0 + 0.349 207 447 240 704;
  • 16) 0.349 207 447 240 704 × 2 = 0 + 0.698 414 894 481 408;
  • 17) 0.698 414 894 481 408 × 2 = 1 + 0.396 829 788 962 816;
  • 18) 0.396 829 788 962 816 × 2 = 0 + 0.793 659 577 925 632;
  • 19) 0.793 659 577 925 632 × 2 = 1 + 0.587 319 155 851 264;
  • 20) 0.587 319 155 851 264 × 2 = 1 + 0.174 638 311 702 528;
  • 21) 0.174 638 311 702 528 × 2 = 0 + 0.349 276 623 405 056;
  • 22) 0.349 276 623 405 056 × 2 = 0 + 0.698 553 246 810 112;
  • 23) 0.698 553 246 810 112 × 2 = 1 + 0.397 106 493 620 224;
  • 24) 0.397 106 493 620 224 × 2 = 0 + 0.794 212 987 240 448;
  • 25) 0.794 212 987 240 448 × 2 = 1 + 0.588 425 974 480 896;
  • 26) 0.588 425 974 480 896 × 2 = 1 + 0.176 851 948 961 792;
  • 27) 0.176 851 948 961 792 × 2 = 0 + 0.353 703 897 923 584;
  • 28) 0.353 703 897 923 584 × 2 = 0 + 0.707 407 795 847 168;
  • 29) 0.707 407 795 847 168 × 2 = 1 + 0.414 815 591 694 336;
  • 30) 0.414 815 591 694 336 × 2 = 0 + 0.829 631 183 388 672;
  • 31) 0.829 631 183 388 672 × 2 = 1 + 0.659 262 366 777 344;
  • 32) 0.659 262 366 777 344 × 2 = 1 + 0.318 524 733 554 688;
  • 33) 0.318 524 733 554 688 × 2 = 0 + 0.637 049 467 109 376;
  • 34) 0.637 049 467 109 376 × 2 = 1 + 0.274 098 934 218 752;
  • 35) 0.274 098 934 218 752 × 2 = 0 + 0.548 197 868 437 504;
  • 36) 0.548 197 868 437 504 × 2 = 1 + 0.096 395 736 875 008;
  • 37) 0.096 395 736 875 008 × 2 = 0 + 0.192 791 473 750 016;
  • 38) 0.192 791 473 750 016 × 2 = 0 + 0.385 582 947 500 032;
  • 39) 0.385 582 947 500 032 × 2 = 0 + 0.771 165 895 000 064;
  • 40) 0.771 165 895 000 064 × 2 = 1 + 0.542 331 790 000 128;
  • 41) 0.542 331 790 000 128 × 2 = 1 + 0.084 663 580 000 256;
  • 42) 0.084 663 580 000 256 × 2 = 0 + 0.169 327 160 000 512;
  • 43) 0.169 327 160 000 512 × 2 = 0 + 0.338 654 320 001 024;
  • 44) 0.338 654 320 001 024 × 2 = 0 + 0.677 308 640 002 048;
  • 45) 0.677 308 640 002 048 × 2 = 1 + 0.354 617 280 004 096;
  • 46) 0.354 617 280 004 096 × 2 = 0 + 0.709 234 560 008 192;
  • 47) 0.709 234 560 008 192 × 2 = 1 + 0.418 469 120 016 384;
  • 48) 0.418 469 120 016 384 × 2 = 0 + 0.836 938 240 032 768;
  • 49) 0.836 938 240 032 768 × 2 = 1 + 0.673 876 480 065 536;
  • 50) 0.673 876 480 065 536 × 2 = 1 + 0.347 752 960 131 072;
  • 51) 0.347 752 960 131 072 × 2 = 0 + 0.695 505 920 262 144;
  • 52) 0.695 505 920 262 144 × 2 = 1 + 0.391 011 840 524 288;
  • 53) 0.391 011 840 524 288 × 2 = 0 + 0.782 023 681 048 576;
  • 54) 0.782 023 681 048 576 × 2 = 1 + 0.564 047 362 097 152;
  • 55) 0.564 047 362 097 152 × 2 = 1 + 0.128 094 724 194 304;
  • 56) 0.128 094 724 194 304 × 2 = 0 + 0.256 189 448 388 608;
  • 57) 0.256 189 448 388 608 × 2 = 0 + 0.512 378 896 777 216;
  • 58) 0.512 378 896 777 216 × 2 = 1 + 0.024 757 793 554 432;
  • 59) 0.024 757 793 554 432 × 2 = 0 + 0.049 515 587 108 864;
  • 60) 0.049 515 587 108 864 × 2 = 0 + 0.099 031 174 217 728;
  • 61) 0.099 031 174 217 728 × 2 = 0 + 0.198 062 348 435 456;
  • 62) 0.198 062 348 435 456 × 2 = 0 + 0.396 124 696 870 912;
  • 63) 0.396 124 696 870 912 × 2 = 0 + 0.792 249 393 741 824;
  • 64) 0.792 249 393 741 824 × 2 = 1 + 0.584 498 787 483 648;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 254 797 590 553(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001(2)

6. Positive number before normalization:

0.000 254 797 590 553(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 254 797 590 553(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001 =

0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001


Decimal number -0.000 254 797 590 553 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 1000 1010 1101 0110 0100 0001

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100